A) \[+7.3\text{ }kcal\]
B) more than 7.3 kcal
C) less than 7.3 kcal
D) zero
Correct Answer: C
Solution :
\[\Delta H=\Delta E+\Delta {{n}_{g}}RT\] Or \[\Delta E=\Delta H-\Delta {{n}_{g}}RT\] \[\Delta {{n}_{g}}=\frac{1}{2}\] \[\Delta E=2.3\times {{10}^{3}}-\frac{1}{2}\times 2\times 298\] \[=7300\text{ }cal-298\] \[=7002\,cal\] \[\approx 7.00\,kcal\]You need to login to perform this action.
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